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Study Guides > Intermediate Algebra

Read: Define Solutions to Systems of Linear Inequalities

Learning Objectives

  • Graph a system of linear inequalities and define the solutions region
  • Verify whether a point is a solution to a system of inequalities
  • Identify when a system of inequalities has no solution
 

Graph a system of two inequalities

Remember from the module on graphing that the graph of a single linear inequality splits the coordinate plane into two regions. On one side lie all the solutions to the inequality. On the other side, there are no solutions. Consider the graph of the inequality [latex]y<2x+5[/latex]. An upward-sloping dotted line with the region below it shaded. The shaded region is labeled y is less than 2x+5. A is equal to (-1,1). B is equal to (3,1). The dashed line is [latex]y=2x+5[/latex]. Every ordered pair in the shaded area below the line is a solution to [latex]y<2x+5[/latex], as all of the points below the line will make the inequality true. If you doubt that, try substituting the x and y coordinates of Points A and B into the inequality—you’ll see that they work. So, the shaded area shows all of the solutions for this inequality. The boundary line divides the coordinate plane in half. In this case, it is shown as a dashed line as the points on the line don’t satisfy the inequality. If the inequality had been [latex]y\leq2x+5[/latex], then the boundary line would have been solid. Let’s graph another inequality: [latex]y>−x[/latex]. You can check a couple of points to determine which side of the boundary line to shade. Checking points M and N yield true statements. So, we shade the area above the line. The line is dashed as points on the line are not true. Downward-sloping dotted line with the region above it shaded. The shaded region is y is greater than negative x. Point M=(-2,3). Point N=(4,-1). To create a system of inequalities, you need to graph two or more inequalities together. Let’s use [latex]y<2x+5[/latex] and [latex]y>−x[/latex] since we have already graphed each of them. The two previous graphs combined. A blue dotted line with the region above shaded and labeled y is greater than negative x. A red dotted line with the region below it shaded and labeled y is less than 2x+5. The region where the shaded areas overlap is labeled y is greater than negative x and y is less than 2x+5. The point M equals (-2,3) and is in the blue shaded region. The point A equals (-1,-1) and is in the red shaded region. The point B equals (3,1) and is in the purple overlapping region. The point N equals (4,-1) and is also in the purple overlapping region. The purple area shows where the solutions of the two inequalities overlap. This area is the solution to the system of inequalities. Any point within this purple region will be true for both [latex]y>−x[/latex] and [latex]y<2x+5[/latex]. In the following video examples, we show how to graph a system of linear inequalities, and define the solution region. https://youtu.be/ACTxJv1h2_c https://youtu.be/cclH2h1NurM In the next section, we will see that points can be solutions to systems of equations and inequalities.  We will verify algebraically whether a point is a solution to a linear equation or inequality.

Determine whether an ordered pair is a solution to a system of linear inequalities

image014 On the graph above, you can see that the points B and N are solutions for the system because their coordinates will make both inequalities true statements. In contrast, points M and A both lie outside the solution region (purple). While point M is a solution for the inequality [latex]y>−x[/latex] and point A is a solution for the inequality [latex]y<2x+5[/latex], neither point is a solution for the system. The following example shows how to test a point to see whether it is a solution to a system of inequalities.

Example

Is the point [latex](2, 1)[/latex] a solution of the system [latex]x+y>1[/latex] and [latex]2x+y<8[/latex]?

Answer: Check the point with each of the inequalities. Substitute [latex]2[/latex]for x and [latex]1[/latex] for y. Is the point a solution of both inequalities? [latex-display]\begin{array}{r}x+y>1\\2+1>1\\3>1\\\text{TRUE}\end{array}[/latex-display] [latex](2, 1)[/latex] is a solution for [latex]x+y>1[/latex]. [latex-display]\begin{array}{r}2x+y<8\\2\left(2\right)+1<8\\4+1<8\\5<8\\\text{TRUE}\end{array}[/latex-display] [latex-display](2, 1)[/latex] is a solution for [latex]2x+y<8.[/latex-display] Since [latex](2, 1)[/latex] is a solution of each inequality, it is also a solution of the system.

Answer

The point [latex](2, 1)[/latex] is a solution of the system [latex]x+y>1[/latex] and [latex]2x+y<8[/latex].

Here is a graph of the system in the example above. Notice that [latex](2, 1)[/latex] lies in the purple area, which is the overlapping area for the two inequalities. Two dotted lines, one red and one blue. The region below the blue dotted line is shaded and labeled 2x+y is less than 8. The region above the dotted red line is shaded and labeled x+y is greater than 1. The overlapping shaded region is purple and is labeled x+y is greater than 1 and 2x+y is less than 8. The point (2,1) is in the overlapping purple region.

Example

Is the point [latex](2, 1)[/latex] a solution of the system [latex]x+y>1[/latex] and [latex]3x+y<4[/latex]?

Answer: Check the point with each of the inequalities. Substitute [latex]2[/latex] for x and [latex]1[/latex] for y. Is the point a solution of both inequalities? [latex-display]\begin{array}{r}x+y>1\\2+1>1\\3>1\\\text{TRUE}\end{array}[/latex-display] [latex](2, 1)[/latex] is a solution for [latex]x+y>1[/latex]. [latex-display]\begin{array}{r}3x+y<4\\3\left(2\right)+1<4\\6+1<4\\7<4\\\text{FALSE}\end{array}[/latex-display] [latex](2, 1)[/latex] is not a solution for [latex]3x+y<4[/latex]. Since [latex](2, 1)[/latex] is not a solution of one of the inequalities, it is not a solution of the system.

Answer

The point [latex](2, 1)[/latex] is not a solution of the system [latex]x+y>1[/latex] and [latex]3x+y<4[/latex].

Here is a graph of this system. Notice that [latex](2, 1)[/latex] is not in the purple area, which is the overlapping area; it is a solution for one inequality (the red region), but it is not a solution for the second inequality (the blue region). A downward-sloping bue dotted line with the region below shaded and labeled 3x+y is less than 4. A downward-sloping red dotted line with the region above it shaded and labeled x+y is greater than 1. An overlapping purple shaded region is labeled x+y is greater than 1 and 3x+y is less than 4. A point (2,1) is in the red shaded region, but not the blue or overlapping purple shaded region. In the following video we show another example of determining whether a point is in the solution of a system of linear inequalities. https://youtu.be/o9hTFJEBcXs As shown above, finding the solutions of a system of inequalities can be done by graphing each inequality and identifying the region they share. Below, you are given more examples that show the entire process of defining the region of solutions on a graph for a system of two linear inequalities.  The general steps are outlined below:
  • Graph each inequality as a line and determine whether it will be solid or dashed
  • Determine which side of each boundary line represents solutions to the inequality by testing a point on each side
  • Shade the region that represents solutions for both inequalities

Systems with no solutions

In the next example, we will show the solution to a system of two inequalities whose boundary lines are parallel to each other.  When the graphs of a system of two linear equations are parallel to each other, we found that there was no solution to the system.  We will get a similar result for the following system of linear inequalities.

Examples

Graph the system [latex]\begin{array}{c}y\ge2x+1\\y\lt2x-3\end{array}[/latex]

Answer: The boundary lines for this system are parallel to each other, note how they have the same slopes.

[latex]\begin{array}{c}y=2x+1\\y=2x-3\end{array}[/latex]

Plotting the boundary lines will give the graph below. Note that the inequality [latex]y\lt2x-3[/latex] requires that we draw a dashed line, while the inequality [latex]y\ge2x+1[/latex] will require a solid line. y=2x+1 Now we need to add the regions that represent the inequalities.  For the inequality [latex]y\ge2x+1[/latex] we can test a point on either side of the line to see which region to shade. Let's test [latex]\left(0,0\right)[/latex] to make it easy. Substitute [latex]\left(0,0\right)[/latex] into [latex]y\ge2x+1[/latex]

[latex]\begin{array}{c}y\ge2x+1\\0\ge2\left(0\right)+1\\0\ge{1}\end{array}[/latex]

This is not true, so we know that we need to shade the other side of the boundary line for the inequality  [latex]y\ge2x+1[/latex]. The graph will now look like this: y=2x+1 Now let's shade the region that shows the solutions to the inequality [latex]y\lt2x-3[/latex].  Again, we can pick [latex]\left(0,0\right)[/latex] to test because it makes easy algebra. Substitute [latex]\left(0,0\right)[/latex] into [latex]y\lt2x-3[/latex]

[latex]\begin{array}{c}y\lt2x-3\\0\lt2\left(0,\right)x-3\\0\lt{-3}\end{array}[/latex]

This is not true, so we know that we need to shade the other side of the boundary line for the inequality[latex]y\lt2x-3[/latex]. The graph will now look like this: y=2x+1 This system of inequalities shares no points in common.  

 

Summary

  • Solutions to systems of linear inequalities are entire regions of points
  • You can verify whether a point is a solution to a system of linear inequalities in the same way you verify whether a point is a solution to a system of equations
  • Systems of inequalities can have no solutions when boundary lines are parallel

Licenses & Attributions

CC licensed content, Shared previously

  • Ex 1: Graph a System of Linear Inequalities. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Ex 2: Graph a System of Linear Inequalities. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
  • Unit 14: Systems of Equations and Inequalities, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology Located at: https://www.nroc.org/. License: CC BY: Attribution.